Given fx x25 x0 find its inverse if possible a f1x square ro

Given f(x)= x25, x0, find its inverse if possible.

a) f1(x)= square root 2x+5

b) f1(x)= square root x5

c) f1(x)= square root x+5

d) f1(x)= square root x+5

e) f1(x)= square root x5

Solution

f(x) = x^2 - 5

vertex is at ( 0, -5) for x<=0 ; f(x) is 1 to1 function so inverse exists

To find inverse : Plug y = x and x=y , solve for y :

y = x^2 - 5

x = y^2 - 5

x+5 = y^2

y = + / - sqrt(x+5)

for domain x<=0 ; f^-1(x) = -sqrt(x+5)

Option d)